To solve the inequality [tex]\( |2x + 1| > -x^2 + 4 \)[/tex], we need to carefully analyze and break down the components of the inequality.
### Step 1: Understand the Absolute Value Expression
The absolute value [tex]\( |2x + 1| \)[/tex] represents a piecewise function. We can define it as follows:
[tex]\[
|2x + 1| =
\begin{cases}
2x + 1 & \text{if } 2x + 1 \geq 0 \\
-(2x + 1) & \text{if } 2x + 1 < 0
\end{cases}
\][/tex]
Solving [tex]\( 2x + 1 \geq 0 \)[/tex] gives us [tex]\( x \geq -\frac{1}{2} \)[/tex]. Therefore:
[tex]\[
|2x + 1| =
\begin{cases}
2x + 1 & \text{if } x \geq -\frac{1}{2} \\
-2x - 1 & \text{if } x < -\frac{1}{2}
\end{cases}
\][/tex]
### Step 2: Set up the Inequalities
Now we solve the inequality [tex]\( |2x + 1| > -x^2 + 4 \)[/tex] for each case.
#### Case 1: When [tex]\( x \geq -\frac{1}{2} \)[/tex]
Here, [tex]\( |2x + 1| = 2x + 1 \)[/tex]. The inequality becomes:
[tex]\[
2x + 1 > -x^2 + 4
\][/tex]
Rearrange to standard form:
[tex]\[
x^2 + 2x + 1 > 0
\][/tex]
Factorize:
[tex]\[
(x + 1)^2 > 0
\][/tex]
The quadratic [tex]\( (x + 1)^2 \)[/tex] is always non-negative, and greater than 0 for all [tex]\( x \neq -1 \)[/tex]. Thus:
[tex]\[
x \in (-\infty, -1) \cup (-1, \infty)
\][/tex]
However, since we are focusing on [tex]\( x \geq -\frac{1}{2} \)[/tex] in this case, this solution simplifies to:
[tex]\[
x \in [-\frac{1}{2}, -1) \cup (-1, \infty)
\][/tex]
#### Case 2: When [tex]\( x < -\frac{1}{2} \)[/tex]
Here, [tex]\( |2x + 1| = -2x - 1 \)[/tex]. The inequality becomes:
[tex]\[
-2x - 1 > -x^2 + 4
\][/tex]
Rearrange to standard form:
[tex]\[
x^2 - 2x - 5 > 0
\][/tex]
Solve the quadratic inequality by finding the roots:
[tex]\[
x^2 - 2x - 5 = 0
\][/tex]
Using the quadratic formula:
[tex]\[
x = \frac{2 \pm \sqrt{4 + 20}}{2} = 1 \pm \sqrt{6}
\][/tex]
So the roots are [tex]\( x = 1 + \sqrt{6} \)[/tex] and [tex]\( x = 1 - \sqrt{6} \)[/tex]. For [tex]\( x < -\frac{1}{2} \)[/tex], we want the intervals where the quadratic is positive. The quadratic opens upwards, so:
[tex]\[
x \in (-\infty, 1 - \sqrt{6}) \cup (1 + \sqrt{6}, \infty)
\][/tex]
Restricting to [tex]\( x < -\frac{1}{2} \)[/tex]:
[tex]\[
x \in (-\infty, 1 - \sqrt{6})
\][/tex]
### Step 3: Combine the Solution Intervals
From Case 1, we have:
[tex]\[
x \in [-\frac{1}{2}, -1) \cup (-1, \infty)
\][/tex]
From Case 2, we have:
[tex]\[
x \in (-\infty, 1 - \sqrt{6})
\][/tex]
Combining and simplifying these intervals, we get the final solution as:
[tex]\[
x \in (-\infty, 1 - \sqrt{6}) \cup (1, \infty)
\][/tex]
